Question

A circular loop of wire with radius r = 0.0250 m and resistance R = 0.390 Ω is in a region of the spatially uniform magnetic field, as shown in Fig. E29.23. The magnetic field is directed into the plane of the figure. At t = 0, B = 0. The magnetic field then begins increasing, with B(t) = $\left(0.387T/s^3\right)t^3$. What is the current in the loop (magnitude and direction) at the instant when B = 1.33 T?

Short answer

answer is not given in the drive

Step-by-step solution

Step 1:

Our target variable is the current in the loop at the instant B = 1.33 T. The direction of

the current is specified by Lenz's law. The magnitude of the current in the loop with

resistance R is given by

I= $\epsilon /R$

Where $\epsilon$, the induced emf, is given by Faraday's

law. We need to solve for the time at which B =1.33 T to evaluate

the induced emf and thus the current at this time.

By Lenz's law, the induced current must produce a magnetic field$B_{induced}$inside the loop that is upward, opposing the change in flux from the right-hand rule we described for the direction of the magnetic field produced by a circular loop $B_{induced}$, which will be in the desired direction if the induced current flows as shown in Fig. 1 in the

Counter clock-wise direction. The magnetic field is uniform over the loop and perpendicular to the plane of the loop, so the magnetic flux $B_{induced}=BA,A=\pi r^2$is the area of the circular loop. Hence, by Faraday's law the induced a EMFs.

$\epsilon =-\frac{d{\Phi }_B}{dt}=-\frac{d}{dt}\left(BA\right)=-A\frac{dB}{dt}=-\pi r^2\frac{dB}{dt}$

So, the magnitude of the current in the loop is

$I=\frac{\left|\epsilon \right|}{R}=\frac{\pi r^2}{R}\left|\frac{dB}{dt}\right|.$

The time derivative of B is $dB/dt=(1.14T/s^3)t^2$so, we obtain

$I=\frac{\pi r^2}{R}(1.14T/s^3)t^2$

The time at a particular field.

The time t* corresponding to the field B = 1.33T satisfies the equation B(t*) =1.33T. Solving for t* gives

$$\begin{gathered} (0.380T/s^3)(t^{*}{)}^3=1.33T \\ {t}^{*}={}^a\sqrt{\frac{1.33T}{0.380T/s^3}}=1.52s \end{gathered}$$

Substitute this time and the known values of r and R into the expression for I gives

$I=\frac{\pi (0.0250m{)}^2}{0.390\Omega }(1.14T/s^3)(1.52s{)}^2=13.3mA$

Hence, the value of I is 13.3mA.